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Let $H$ be Hilbert space, $\{a_n\}_{n=1},$ be ONS and $K$ be a compact operator. Suppose $K_nx:=(Kx,a_n)a_n$, $\sum _{n=1}^{N} K_n$ convergent as $N \to \infty$. So,$||K_nx||=|(Kx,a_n)|=|(x,K^* a_n)| \leq ||x|| ||K^* a_n||$ and so $\lim ||K_n||\leq \lim ||K^* a_n||=0$ since $K^*$ is compact and $\{a_n\}$ weakly convergent.

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Let $x$ be an element of the closed unit ball. Then by orthogonality of $\left(b_n\right)$, $$ \left\lVert \sum_{n=M}^N K_nx\right\rVert^2=\sum_{n=M}^N \left(Kx,a_n\right)^2. $$ Since the set $\left\{Kx,x\in H,\left\lVert x\right\rVert\leqslant 1\right\}$ has a compact closure, it suffices to prove that for each subset $C$ of $H$ having a compact closure,
$$ \lim_{M,N\to +\infty}\sup_{y\in C}\sum_{n=M}^N \left(y,a_n\right)^2=0. $$ This is clear when $C$ is finite; use precompactness to generalize.

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Let $L : H\to H$ be given by

$$ L(x) = \sum_{n=1}^\infty \langle x, a_n\rangle b_n.$$

Then

$$K_N :=\sum_{n=1}^N K_n =L_N \circ K,$$

where $L_n (y)= \sum_{n=1}^N \langle y, a_n \rangle b_n.$ To show that $K_N$ converges, it suffices to show that $L^{-1}\circ K_N$ converges to $K$.But

$$L^{-1} \circ K_N (x)= \sum_{n=1}^N \langle Kx, a_n\rangle a_n$$

and it reduces to the usual argument where one proves that finite rank operators in Hilbert space are dense in the space of compact operators, that can be found here.

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